On the one-side equivalence of matrices with given canonical diagonal form

On the one-side equivalence of matrices with given canonical diagonal form

The simpler form of a matrix with canonical diagonal form diag(1,…,1,φ,…,φ) obtained by the one-side transformation is determined.

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Дата:2011
Автор: Shchedryk, V.
Формат: Стаття
Мова:English
Опубліковано: Інститут прикладної математики і механіки НАН України 2011
Назва видання:Algebra and Discrete Mathematics
Онлайн доступ:http://dspace.nbuv.gov.ua/handle/123456789/154769
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Назва журналу:Digital Library of Periodicals of National Academy of Sciences of Ukraine
Цитувати:On the one-side equivalence of matrices with given canonical diagonal form / V. Shchedryk // Algebra and Discrete Mathematics. — 2011. — Vol. 12, № 2. — С. 102–111. — Бібліогр.: 5 назв. — англ.

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Digital Library of Periodicals of National Academy of Sciences of Ukraine
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spelling irk-123456789-1547692019-06-16T01:31:27Z On the one-side equivalence of matrices with given canonical diagonal form Shchedryk, V. The simpler form of a matrix with canonical diagonal form diag(1,…,1,φ,…,φ) obtained by the one-side transformation is determined. 2011 Article On the one-side equivalence of matrices with given canonical diagonal form / V. Shchedryk // Algebra and Discrete Mathematics. — 2011. — Vol. 12, № 2. — С. 102–111. — Бібліогр.: 5 назв. — англ. 1726-3255 2000 Mathematics Subject Classification:15A21. http://dspace.nbuv.gov.ua/handle/123456789/154769 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України
institution Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection DSpace DC
language English
description The simpler form of a matrix with canonical diagonal form diag(1,…,1,φ,…,φ) obtained by the one-side transformation is determined.
format Article
author Shchedryk, V.
spellingShingle Shchedryk, V.
On the one-side equivalence of matrices with given canonical diagonal form
Algebra and Discrete Mathematics
author_facet Shchedryk, V.
author_sort Shchedryk, V.
title On the one-side equivalence of matrices with given canonical diagonal form
title_short On the one-side equivalence of matrices with given canonical diagonal form
title_full On the one-side equivalence of matrices with given canonical diagonal form
title_fullStr On the one-side equivalence of matrices with given canonical diagonal form
title_full_unstemmed On the one-side equivalence of matrices with given canonical diagonal form
title_sort on the one-side equivalence of matrices with given canonical diagonal form
publisher Інститут прикладної математики і механіки НАН України
publishDate 2011
url http://dspace.nbuv.gov.ua/handle/123456789/154769
citation_txt On the one-side equivalence of matrices with given canonical diagonal form / V. Shchedryk // Algebra and Discrete Mathematics. — 2011. — Vol. 12, № 2. — С. 102–111. — Бібліогр.: 5 назв. — англ.
series Algebra and Discrete Mathematics
work_keys_str_mv AT shchedrykv ontheonesideequivalenceofmatriceswithgivencanonicaldiagonalform
first_indexed 2025-07-14T06:52:42Z
last_indexed 2025-07-14T06:52:42Z
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fulltext Algebra and Discrete Mathematics RESEARCH ARTICLE Volume 12 (2011). Number 2. pp. 102 – 111 c© Journal “Algebra and Discrete Mathematics” On the one-side equivalence of matrices with given canonical diagonal form Volodymyr Shchedryk Communicated by M. Ya. Komarnytskyj Abstract. The simpler form of a matrix with canonical diagonal form diag(1, . . . , 1, ϕ, . . . , ϕ) obtained by the one-side trans- formation is determined. Let R be an adequate ring [1] i.e. a commutative domain in which every finitely generated ideal is principal, and which further satisfies the following condition: for any a, c ∈ R with a 6= 0 , one can write a = rs with (r, c) = 1 and (s′, c) 6= 1 for any non unit divisor s′ of s. Let A be an n× n matrix over R. It is known [1] that there exist invertible matrices P,Q, such that PAQ = diag(ϕ1, . . . , ϕn) = Φ. (1) The matrix Φ is called the canonical diagonal form of the matrix A, ϕi| ϕi+1, i = 1, . . . , n− 1. In solving of some matrix problems especially factorization of matrices [2,3], in description of all the Abelian subgroups [4], there emerges the necessity of finding all the non-associated matrices with canonical diagonal form given beforehand. Usual Hermite normal form does not approach to our purposes because it evaluates in the rough way and gives a possibility to describe non-associated matrices with set-up determinant only. That is why there emerges the necessity of building such form of matrix with respect to one sided transformation, giving a glance to which is enough to make a decision as for its canonical diagonal form. The equality (1) gives us a possibility to write matrix A in the following way A = P−1ΦQ−1. Making changes in its right part we will have a new form 2000 Mathematics Subject Classification: 15A21. Key words and phrases: adequate ring, canonical diagonal form, Hermite normal form, one-side equivalence of matrices, invariants, primitive matrices. V. P. Shchedryk 103 P−1Φ. But this type of matrices is not a normal form of the matrix A as for the right side changes because the matrix P determined ambiguously. By [2] the set PA of all invertible matrices which satisfies equation (1) has the form PA = GΦP, where GΦ = {H ∈ GLn(R) | HΦ = ΦH1, H1 ∈ GLn(R)} . This set is a multiplicative group and if detΦ 6= 0 consists of all invertible matrices of the form H = ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ h11 h12 ... h1.n−1 h1n ϕ2 ϕ1 h21 h22 ... h2.n−1 h2n ... ... ... ... ... ϕn ϕ1 hn1 ϕn ϕ2 hn2 ... ϕn ϕn−1 hn.n−1 hnn ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ . Thus, PA is a left conjugacy class GLn(R) with respect to the group GΦ. Therefore, in order that the matrix P−1Φ be a normal form of the matrix A, with respect to the transformation from the right, it is necessary either to choose a representative in the class GΦP or, what is the same, indicate the normal form of the invertible matrices with respect to the action of the group GΦ. The present paper is devoted to the investigation of this question. Let Φ = Et ⊕ ϕEn−t, Φ∗ = ϕEt ⊕ En−t, ϕ 6= 0, 1 ≤ t < n, where Et is the identity t× t matrix. In this case, the group GΦ consists of all invertible matrices of the form ∥ ∥ ∥ ∥ H11 H12 ϕH21 H22 ∥ ∥ ∥ ∥ , where H11 is a t× t matrix. A matrix is called primitive if the greatest common divisor of minor of maximal order is equal to 1. The matrix A is called left associate to the matrix B if A = UB, where U ∈ GLn(R). This fact will be denoted A l ∼B. Lemma 1. Let B = ∥ ∥ ∥ ∥ ∥ ∥ B1 B2 B3 ∥ ∥ ∥ ∥ ∥ ∥ be a primitive n× (n− k + 1) matrix, t < k < n. The matrices B1, B3 is t× (n− k + 1), (n− k + 1)× (n− k + 1) matrices, respectively. Let Φ∗B l ∼ ∥ ∥ ∥ ∥ ∥ ∥ 0 0 B3 ∥ ∥ ∥ ∥ ∥ ∥ . (2) 104 On the one-side equivalence of matrices Then there exists a matrix H ∈ GΦ such that HB = ∥ ∥ ∥ ∥ ∥ ∥ B1 0 B3. ∥ ∥ ∥ ∥ ∥ ∥ . Proof. Consider the matrix equation XB3 = ∥ ∥ ∥ ∥ ϕB1 B2 ∥ ∥ ∥ ∥ . (3) The matrix Φ∗B = ∥ ∥ ∥ ∥ ∥ ∥ ϕB1 B2 B3 ∥ ∥ ∥ ∥ ∥ ∥ is extended matrix of equation (3). From (2) it follows that the invariant factors of the matrices Φ∗B,B3 are equal. By Theorem 2 from [3, p. 218] equation (3) has the solution X = U = ∥ ∥ ∥ ∥ U1 U2 ∥ ∥ ∥ ∥ , where U1 is a t×(n−k+1) matrix and U2 is a (k − t− 1)× (n− k + 1) matrix. Then ∥ ∥ ∥ ∥ ∥ ∥ Et 0 −U1 0 Ek−t−1 −U2 0 0 En−k+1 ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ϕB1 B2 B3 ∥ ∥ ∥ ∥ ∥ ∥ = = ∥ ∥ ∥ ∥ Ek−1 −U 0 En−k+1 ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ϕB1 B2 B3 ∥ ∥ ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ ∥ ∥ 0 0 B3 ∥ ∥ ∥ ∥ ∥ ∥ . This implies that ∥ ∥ ∥ ∥ ∥ ∥ Et 0 0 0 Ek−t−1 −U2 0 0 En−k+1 ∥ ∥ ∥ ∥ ∥ ∥ ︸ ︷︷ ︸ H ∥ ∥ ∥ ∥ ∥ ∥ B1 B2 B3 ∥ ∥ ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ ∥ ∥ B1 0 B3 ∥ ∥ ∥ ∥ ∥ ∥ . Observing that H ∈ GΦ, we conclude the proof of the lemma. Lemma 2. Let A be an n×m matrix and H ∈ GΦ. Then Φ∗HA l ∼Φ∗A. Proof. Since H = ∥ ∥ ∥ ∥ H11 H12 ϕH21 H22 ∥ ∥ ∥ ∥ , V. P. Shchedryk 105 where H11 is a t× t matrix we have Φ∗H = ∥ ∥ ∥ ∥ ϕH11 ϕH12 ϕH21 H22 ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ H11 ϕH12 H21 H22 ∥ ∥ ∥ ∥ Φ∗ = H1Φ∗. The matrix Φ∗ is nonsingular, so that detH = detH1 i.e. the matrix H1 is invertible. Consequently, Φ∗HA = H1Φ∗A l ∼Φ∗A. Lemma 3. Let Bk = ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ b11 b12 b13 . . . b1.n−k b1.n−k+1 . . . . . . . . . . . . . . . . . . bt1 bt2 bt3 . . . bt.n−k bt.n−k+1 bt+1.1 0 0 . . . 0 0 . . . . . . . . . . . . . . . . . . bk1 0 0 . . . 0 0 bk+1.1 βk+1 0 0 0 bk+2.1 bk+2.2 βk+2 0 0 ... ... . . . bn−1.1 bn−1.2 bn−1.3 βn−1 0 bn1 bn2 bn3 . . . bn.n−k βn ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ = = ∥ ∥ ∥ ∥ ∥ ∥ B11 B12 B21 0 B31 B32 ∥ ∥ ∥ ∥ ∥ ∥ be a primitive n× (n− k + 1) matrix, t < k < n, and Φ∗Bk l ∼ ∥ ∥ ∥ ∥ ∥ ∥ 0 0 D21 0 B31 B32 ∥ ∥ ∥ ∥ ∥ ∥ , (4) where D21 = ∥ ∥ 0 . . . 0 βk ∥ ∥T . Then there exists a matrix H ∈ GΦ such that HBk = ∥ ∥ ∥ ∥ ∥ ∥ B′ 11 B′ 12 D21 0 B31 B32 ∥ ∥ ∥ ∥ ∥ ∥ . (5) 106 On the one-side equivalence of matrices Proof. Consider the equation XB32 = ϕB12. (6) The equality Φ∗Bk = ∥ ∥ ∥ ∥ ∥ ∥ ϕB11 ϕB12 B21 0 B31 B32 ∥ ∥ ∥ ∥ ∥ ∥ , is valid. From (4) we conclude that ∥ ∥ ∥ ∥ ∥ ∥ ϕB12 0 B32 ∥ ∥ ∥ ∥ ∥ ∥ l ∼ ∥ ∥ ∥ ∥ ∥ ∥ 0 0 B32 ∥ ∥ ∥ ∥ ∥ ∥ . This implies that the invariant factors of the matrices B32, ∥ ∥ ∥ ∥ ϕB12 B32 ∥ ∥ ∥ ∥ are equal. By Theorem 2 from [3, p. 218], equation (6) has the solution X = U13. Thus, the equality ∥ ∥ ∥ ∥ ∥ ∥ Et 0 −U13 0 Ek−t 0 0 0 En−k ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ϕB11 ϕB12 B21 0 B31 B32 ∥ ∥ ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ ∥ ∥ B′ 11 0 B21 0 B31 B32 ∥ ∥ ∥ ∥ ∥ ∥ , holds, where B′ 11 = ∥ ∥ b′11 . . . b′t1 ∥ ∥T . By Lemma 2, ∥ ∥ ∥ ∥ ∥ ∥ B′ 11 B′ 12 B21 0 B31 B32 ∥ ∥ ∥ ∥ ∥ ∥ l ∼ ∥ ∥ ∥ ∥ ∥ ∥ 0 0 D21 0 B31 B32 ∥ ∥ ∥ ∥ ∥ ∥ , so that (b′11, . . . , b ′ t1, bt+1.1, . . . , bk1) = βk. According to property 6 from [5], there exist v1, . . . , vk such that v1b ′ 11 + · · ·+ vtb ′ t1 + vt+1bt+1.1 + · · ·+ vkbk1 = βk, and (vk, ϕ) = 1. Let us complement the primitive row ∥ ∥ v1 . . . vk ∥ ∥ to an invertible matrix Vk in which this row is the last. Consider the invertible matrix ∥ ∥ ∥ ∥ Vk 0 0 En−k ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ Et 0 −U13 0 Ek−t 0 0 0 En−k ∥ ∥ ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ Vk U 0 En−k ∥ ∥ ∥ ∥ = V. V. P. Shchedryk 107 Taking into account that ∥ ∥ v1 . . . vk uk+1 . . . un ∥ ∥ is the k-th row of this matrix, we obtain ∥ ∥ v1 . . . vk uk+1 . . . un ∥ ∥Φ∗Bk = ∥ ∥ βk 0 . . . 0 ∥ ∥ , i.e., ∥ ∥ ϕv1 . . . ϕvt vt+1 . . . vk uk+1 . . . un ∥ ∥Bk = = ∥ ∥ βk 0 . . . 0 ∥ ∥ . Since (v1, . . . , vk) = 1, (vk, ϕ) = 1, we have (ϕv1, . . . , ϕvt, vt+1, . . . , vk) = 1. It means that the matrix Fk = ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ϕv1 . . . ϕvt vt+1 . . . vk uk+1 . . . un 0 . . . 0 0 . . . 0 1 0 . . . . . . . . . . . . . . . . . . . . . 0 . . . 0 0 . . . 0 0 1 ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ is primitive. By property 2 from [5], the matrix Fk can be complemented to an invertible matrix Hk = ∥ ∥ ∥ ∥ ∗ Fk ∥ ∥ ∥ ∥ which belongs to GΦ. Then HkBk = ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ b′11 b′12 . . . b′ 1.n−k+1 . . . . . . . . . . . . b′t1 b′t2 . . . b′t.n−k+1 b′t+1.1 b′t+1.2 . . . b′t+1.n−k+1 . . . . . . . . . . . . b′k−1.1 b′k−1.2 . . . b′k−1.n−k+1 βk 0 0 bk+1.1 βk+1 0 . . . bn1 bn2 βn ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ ∥ ∥ A1 A2 A3 ∥ ∥ ∥ ∥ ∥ ∥ . By Lemma 2 Φ∗HkBk l ∼ ∥ ∥ ∥ ∥ ∥ ∥ 0 0 D21 0 B31 B32 ∥ ∥ ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ ∥ ∥ 0 0 A3 ∥ ∥ ∥ ∥ ∥ ∥ . According to Lemma 1, the group GΦ contain a matrix H ′ k such that H ′ kHk ∥ ∥ ∥ ∥ ∥ ∥ A1 A2 A3 ∥ ∥ ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ ∥ ∥ A1 0 A3 ∥ ∥ ∥ ∥ ∥ ∥ , 108 On the one-side equivalence of matrices which has form (5). The proof is complete. Let us denote by K(f) the set of representatives of the conjugate classes of R/Rf, f ∈ R. Theorem 1. Let B = ‖bij‖ n 1 = ∥ ∥ ∥ ∥ B11 B12 B21 B22 ∥ ∥ ∥ ∥ be an ivertible matrix, where B11 is a t× t matrix and Φ∗B l ∼ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ β1 0 0 ∗ β2 0 . . . ∗ ∗ βn ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ (7) is the left Hermite normal form of the matrix Φ∗B. Then the group GΦ contains a matrix H such that HB = ∥ ∥ ∥ ∥ C11 C12 C21 C22 ∥ ∥ ∥ ∥ , (8) where C22 = ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ βt+1 0 0 ct+2.t+1 βt+2 0 ... . . . cn.t+1 cn.t+2 βn ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ , cij ∈ K(βj), i = t+2, t+3, . . . , n, j = t+1, t+2, . . . , n− 1. The elements cij are uniquely determined and do not depend on the choice of the matrix H. Proof. Using (6) we obtain Φ∗ ∥ ∥ b1n b2n . . . bnn ∥ ∥T ∼ ∥ ∥ 0 . . . 0 βn ∥ ∥T . By Theorem 2 from [6], there exists a matrix Hn ∈ GΦ such that Hn ∥ ∥ b1n b2n . . . bnn ∥ ∥T = ∥ ∥ b′1n . . . b′n−1.n βn ∥ ∥T . According to Lemma 2 Φ∗ ∥ ∥ b′1n . . . b′n−1.n βn ∥ ∥T = = ∥ ∥ ϕb′1n . . . ϕb′tn b′t+1.n . . . b′n−1.n βn ∥ ∥T ∼ ∼ ∥ ∥ 0 . . . 0 βn ∥ ∥T . V. P. Shchedryk 109 Therefore b′in = βndi , i = t+ 1, t+ 2, . . . , n− 1. Then        En−t ⊕ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ 1 0 0 −dt+1 0 1 0 −dt+2 . . . ... 0 0 1 −dn−1 0 0 0 1 ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥        HnB = Bn. Using Lemmas 2 and 3 in consequently to the last two columns of the matrix Bn, to the last three columns of the derived matrix and so fors we get Ht+1 ∈ GΦ such that Ht+1B = ∥ ∥ ∥ ∥ D11 D12 D21 D22 ∥ ∥ ∥ ∥ , where D22 = ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ βt+1 0 0 dt+2.t+1 βt+2 0 ... . . . dn.t+1 dn.t+2 βn ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ . There exists a lower unitriangular matrix U such that UD22 = ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ βt+1 0 0 ct+2.t+1 βt+2 0 ... . . . cn.t+1 cn.t+2 βn ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ is the left Hermite normal form of the matrix D22, i.e., cij ∈ K(βj), i = t+2, t+3, . . . , n, j = t+1, t+2, . . . , n−1 . The the matrix (Et⊕U)Ht+1B has the form (8) and (Et ⊕ U)Ht+1B ∈ GΦ. We will show the uniqueness of the elements cij . Let H1 ∈ GΦ and H1B = ∥ ∥ ∥ ∥ C ′ 11 C ′ 12 C ′ 21 C ′ 22 ∥ ∥ ∥ ∥ , where C ′ 22 = ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ βt+1 0 0 c′t+2.t+1 βt+2 0 ... . . . c′n.t+1 c′n.t+2 βn ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ , c′ij ∈ K(βj) , i = t+ 2, t+ 3, . . . , n, j = t+ 1, t+ 2, . . . , n− 1. Concerning the proof of Lemma 3 we conclude that there exists an upper unitriangular 110 On the one-side equivalence of matrices matrix U such that UΦ∗ ∥ ∥ ∥ ∥ C12 C22 ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ 0 C22 ∥ ∥ ∥ ∥ . (9) Since H ∥ ∥ ∥ ∥ B12 B22 ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ C12 C22 ∥ ∥ ∥ ∥ and H1 ∥ ∥ ∥ ∥ B12 B22 ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ C ′ 12 C ′ 22 ∥ ∥ ∥ ∥ , we obtain ∥ ∥ ∥ ∥ C ′ 12 C ′ 22 ∥ ∥ ∥ ∥ = H2 ∥ ∥ ∥ ∥ C12 C22 ∥ ∥ ∥ ∥ , H2 = H1H −1 ∈ GΦ. Rewrite (9) us UΦ∗H −1 2 H2 ∥ ∥ ∥ ∥ C12 C22 ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ 0 C22 ∥ ∥ ∥ ∥ . By Lemma 2 Φ∗H −1 2 = H3Φ∗, H3 ∈ GLn(R). Thus, UH3Φ∗ ∥ ∥ ∥ ∥ C ′ 12 C ′ 22 ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ 0 C22 ∥ ∥ ∥ ∥ . Let (UH3) −1 = ∥ ∥ ∥ ∥ V11 V12 V21 V22 ∥ ∥ ∥ ∥ , where V11 is a t× t matrix. Then ∥ ∥ ∥ ∥ V11 V12 V21 V22 ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ 0 C22 ∥ ∥ ∥ ∥ = ∥ ∥ ∥ ∥ ϕC ′ 12 C ′ 22 ∥ ∥ ∥ ∥ , i.e., V22C22 = C ′ 22. Since |C22| = |C ′ 22| , the matrix V22 is invertible. Hence, the matrices C22, C ′ 22 are left assosiated. Therefore their left Hermite normal forms are equal. Remark that the matrices C22, C ′ 22 are precisely the left Hermite form. This finished the proof. Theorem 2. Let A = P−1ΦQ−1 , where Φ = Et ⊕ ϕEn−t. Then there exists an invertible matrix U such that AU = V −1Φ, where V is the matrix of the form (8). V. P. Shchedryk 111 Proof. By Theorem 1, there exists H ∈ GΦ such that the matrix HP = V has the form (8). Then A = P−1ΦQ−1 = (HP )−1(HΦ)Q−1 = V −1ΦH1Q −1. Since the matrix H1 is invertible, U = QH−1 1 is desired matrix. References [1] Helmer O. The elementary divisor theorem for certain rings without chain condition, Bull. Amer. Math. Soc., 1943, Vol.49, pp. 225 - 236. [2] Shchedryk V.P. Structure and properties of matrix divisors over commutative elementary divisor domain Math. studii, 1998, Vol.10(2), pp. 115-120. [3] Kazimirskij P.S. Decomposition of matrix polynomials into factors Кyiv, Naukova dumka, 1981. - 224 p. [4] Bhowmik G., Ramare O. Algebra of matrix arithmetic Journal of Algebra, 1998, Vol.210, pp.194 - 215. [5] Shchedryk V.P. Some determinant properties of primitive matrices over Bezout B- domain Algebra and Discrete Mathematics, 2005, №2, pp. 46-57. Contact information Volodymyr Shchedryk Pidstryhach Institute for Applied Problems of mechanics and Mathematics NAS of Ukraine, 3b Naukova Str., L’viv, 79060 E-Mail: shchedrykv@ukr.net Received by the editors: 16.09.2011 and in final form 21.12.2011.

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